O`zbekiston respublikasi oliy va o`rta maxsus ta`lim vazirligi “ “ m m

Mavzu: Ikkinchi va uchinchi tartibli determinantlar.

Determinantlarning asosiy xossalari. Yuqori tartibli determinantlar.

Ikkinchi tartibli kvadrat matritsaga mos keluvchi ikkinchi tartibli determinant

deb quyidagi belgi va tenglik bilan aniqlanuvchi songa aytiladi:

21

12

22

11

22

21

12

11

a

a

a

a

a

a

a

a

×

-

×

=

Uchinchi tartibli kvadrat matritsaga mos keluvchi uchinchi tartibli

determinand deb quyidagi belgi va tenglik bilan aniqlanuvchi songa aytiladi:

11

23

32

33

12

21

13

22

31

33

32

21

31

23

12

33

22

11

33

32

31

23

22

21

13

12

11

a

a

a

a

a

a

a

a

a

a

a

a

a

a

a

a

a

a

a

a

a

a

a

a

a

a

a

-

-

-

+

+

=

Uchinchi tartibli determinantlarni hisoblash uchun  ”uchburchaklar qoidasi”

danfoydalanamiz.

14

·

·

·

·

·

·

·

·

·

-

·

·

·

·

·

·

·

·

·

=

·

·

·

·

·

·

·

·

·

Determinantdagining

ij

a

 elementining

ij

M

 minori deb, bu element turgan qator

va ustunni o`chirish natijasida hosil bo`lgan determinantga aytiladi.

ij

a

 elementining algebraik to`ldiruvchisi deb, musbat yoki manfiy ishora bilan

olingan minorga aytiladi va

( )

ij

j

i

ij

M

A

+

-

= 1

munosabat bilan aniqlanadi.

Ixtiyoriy tartibli determinantni hisoblashning uchta usulini keltiramiz:

1.Determinant tartibini pasaytirish usuli- determinant biror qatori (ustun)

elementlarining bittasidan boshqalarini oldindan nolga aylantirib olib, shu qator

(ustun) bo`yicha yoyish usuli.

Masalan.

( )

91

30

7

13

0

0

30

7

0

13

0

1

2

4

1

32

3

1

15

4

1

2

4

1

1

32

0

3

1

15

0

4

1

2

0

4

8

12

1

3

15

8

2

9

7

3

1

1

23

34

3

5

8

12

1

3

3

=

-

=

-

=

=

-

-

-

=

-

-

=

-

-

-

-

-

-

-

=

A

2. Determinantni uchburchak ko`rinishiga keltirish usuli - determinantning

bosh diagonalidan bir tomonida yotuvchi hamma elementlari nolga aylantiriladi va

uchburchaksimon shaklga keltiriladi, masalan

nn

n

n

a

a

a

a

a

a

...

0

0

...

...

...

...

...

0

...

2

22

1

12

11

=

D

Ravshanki, uchburchak shaklidagi determinantning qiymati bosh diagonallari

elementlari ko`paytmasiga teng:

15

nn

a

a

a

×

×

×

=

D

...

22

11

Masalan.

48

8

3

2

1

8

0

0

0

7

3

0

0

9

5

2

0

4

3

2

1

0

6

4

2

7

3

0

0

9

5

2

0

4

3

2

1

=

×

×

×

=

=

-

-

-

=

D

Determinantni satr yoki ustun bo`yicha yoyib hisoblash quyidagicha bo`ladi:

( )

( )

( )

32

31

22

21

13

3

1

33

31

23

21

12

2

1

33

32

23

22

11

1

1

33

32

31

23

22

21

13

12

11

1

1

1

a

a

a

a

a

a

a

a

a

a

a

a

a

a

a

a

a

a

a

a

a

a

a

a

+

+

+

-

+

-

+

-

=

Masalan.

192

2

1

10

17

8

1

3

4

0

2

1

0

10

17

8

1

3

4

2

4

7

3

1

5

2

2

2

1

6

8

2

8

14

3

2

10

2

1

6

8

0

2

8

14

0

3

2

10

0

0

3

7

1

=

=

-

-

×

=

-

-

-

-

×

×

×

=

-

-

-

-

×

=

-

-

-

-

=

D

3. Sarrius usuli.

.

33

21

12

11

23

32

13

22

31

13

32

21

31

23

12

33

22

11

32

31

22

21

12

11

.

33

32

31

23

22

21

13

12

11

a

a

a

a

a

a

a

a

a

a

a

a

a

a

a

a

a

a

a

a

a

a

a

a

a

a

a

a

a

a

a

a

a

-

-

-

+

+

=

=

D

33

32

31

23

22

21

33

21

12

11

23

32

13

22

31

13

32

21

31

23

12

33

22

11

.

33

32

31

23

22

21

13

12

11

.

a

a

a

a

a

a

a

a

a

a

a

a

a

a

a

a

a

a

a

a

a

a

a

a

a

a

a

a

a

a

a

a

a

-

-

-

+

+

=

=

D

Masalan.

1.

5

47

52

8

9

30

6

10

36

4

1

2

3

1

3

5

3

2

3

1

2

5

1

2

4

3

3

3

5

3

1

2

3

4

3

5

1

3

1

2

2

3

=

-

=

-

-

-

+

+

=

=

×

×

-

×

×

-

×

×

-

×

×

+

×

×

+

×

×

=

2.

5

47

52

8

9

30

6

10

36

1

3

1

2

2

3

4

1

2

3

1

3

5

3

2

3

1

2

5

1

2

4

3

3

4

3

5

1

3

1

2

2

3

=

-

=

-

-

-

+

+

=

=

×

×

-

×

×

-

×

×

-

×

×

+

×

×

+

×

×

=

16

Determinantlarning asosiy xossalari:

a) agar determinantning barcha satrlari mos ustunlari bilan almashtirilsa, uning

qiymati o`zgarmaydi;

b) agar determinant nollardan iborat ustun yoki satrga ega bo`lsa , uning qiymati

nolga teng bo`ladi;

v) agar determinant  ikkita bir xil parallel satr yoki ustunga ega bo`lsa, uning qiymati

nolga teng.

Misollar.

Determinantlarni hisoblang.

1.

3

7

2

5

 2.

4

3

2

1

 3.

5

8

2

3

 4.

12

8

9

6

 5.

2

2

b

ab

ab

a

 6.

1

1

-

+

n

n

n

n

7.

b

a

b

a

b

a

b

a

+

-

-

+

8.

a

a

a

a

cos

sin

sin

cos

-

 9.

b

b

a

a

cos

sin

cos

sin

10.

2

2

2

2

2

2

1

1

1

2

1

2

1

1

t

t

t

t

t

t

t

t

+

-

+

-

+

+

-

11.

x

x

x

x

x

-

-

-

-

1

1

0

1

12.

3

4

1

2

3

5

3

1

2

13.

2

4

3

3

5

2

1

2

3

14.

5

7

2

8

2

3

5

3

4

-

-

-

-

15.

3

2

5

2

1

4

4

2

3

-

-

-

 16.

6

3

1

3

2

1

1

1

1

 17.

0

1

1

1

0

1

1

1

0

 18.

5

0

6

6

1

7

3

0

2

19.

64

8

1

49

7

1

25

5

1

 20.

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

1

-

-

-

Quyidagi determinantlarni ixtiyoriy ustun yoki satr elementlari bo`yicha yoyib

hisoblang.

21.

3

2

1

1

2

5

4

3

2

-

 22.

a

a

a

a

a

1

1

1

1

-

-

 23.

8

4

0

7

5

0

3

2

1

-

24.

b

b

b

b

0

0

0

1

1

 25.

8

4

0

7

5

0

5

2

1

-

 26.

9

8

7

6

5

2

1

0

0

17

27.

8

1

3

7

5

2

6

4

1

-

 28.

3

3

3

2

2

2

1

1

1

 29.

x

x

x

x

x

-

-

-

-

1

1

0

1

30.

10

9

8

7

6

5

2

1

3

-

-

 31.

7

0

4

6

0

2

5

2

1

-

 32.

4

1

1

2

6

2

1

7

1

-

Determinantni tartibini pasaytirish usulidan foydalanib hisoblang:

33.

0

5

2

3

4

1

3

2

3

2

3

4

3

0

4

1

-

-

-

-

-

 34.

2

4

0

3

3

1

2

4

4

2

3

1

5

0

1

2

-

-

-

-

-

-

-

35.

3

5

8

1

2

0

1

5

7

4

1

5

3

0

1

3

-

-

-

-

 36.

3

1

5

0

4

3

7

2

5

4

0

1

2

4

3

6

-

-

-

-

Mavzu:  Chiziqli tenglamalar sistemasini Gauss,Kramer va matritsalar usulida

yechish

1. Ikki noma`lumli ikkita chiziqli tenglamalar sistemasi

î

í

ì

=

+

=

+

2

2

2

1

1

1

c

y

b

x

a

c

y

b

x

a

0

2

2

1

1

¹

=

D

b

a

b

a

 shart bajarilganda

2

2

1

1

2

2

1

1

2

2

1

1

2

2

1

1

,

b

a

b

a

c

a

c

a

y

b

a

b

a

b

c

b

c

x

=

=

Yechimga ega.

Masalan. Ushbu

î

í

ì

=

-

=

+

40

5

4

7

2

3

y

x

y

x

 chiziqli tenglamalar sistemasini yeching.

( )

23

8

15

4

2

5

3

5

4

2

3

-

=

-

-

=

×

-

-

×

=

-

=

D

18

4

23

92

8

15

28

120

5

4

2

3

40

4

7

3

5

23

115

8

15

80

35

5

4

2

3

5

40

2

7

-

=

-

=

-

-

-

=

-

=

=

-

-

=

-

-

-

-

=

-

-

=

y

x

J: (5;-4)

2. Bir jinsli uch noma`lumli ikkita tenglamalar sistemasi

î

í

ì

=

+

+

=

+

+

0

0

2

2

2

1

1

1

z

c

y

b

x

a

z

c

y

b

x

a

ushbu

2

2

1

1

2

2

1

1

2

2

1

1

,

,

b

a

b

a

k

z

c

a

c

a

k

y

c

b

c

b

k

x

=

-

=

=

Formula bilan aniqlanuvchi yechimlarga ega, bunda k- ixtiyoriy son.

Masalan: Ushbu

î

í

ì

=

-

+

=

+

-

0

3

4

0

2

5

2

z

y

x

z

y

x

 tenglamalar sistemesini yeching.

(

)

,

7

8

15

3

4

2

5

k

k

k

x

=

-

=

-

-

=

(

)

,

8

2

6

3

1

2

2

k

k

k

y

-

=

-

-

=

-

=

(

)

.

13

5

8

4

1

5

2

k

k

k

z

=

+

=

-

=

J: x=7k;y=-8k;z=13k.

3. Bir jinsli uch noma`lumli uchta tenglamalar sistemasi berilgan.

ï

î

ï

í

ì

=

+

+

=

+

+

=

+

+

0

0

0

3

3

3

2

2

2

1

1

1

z

c

y

b

x

a

z

c

y

b

x

a

z

c

y

b

x

a

Uning determinanti

0

3

3

3

2

2

2

1

1

1

=

=

D

c

b

a

c

b

a

c

b

a

bo`lsa, tenglamalar sistemasi cheksiz ko`p

yechimga ega.

Misol.Ushbu

ï

î

ï

í

ì

=

+

+

=

-

+

=

+

+

10

2

3

3

3

2

4

3

2

z

y

x

z

y

x

z

y

x

 tenglamalar sistemasini yeching.

19

0

23

23

8

3

9

18

6

2

2

3

3

1

1

2

3

2

1

=

-

=

-

+

-

+

-

=

-

=

D

J: Sistema birgalikda emas.

4. Ikki noma`lumli uchta chiziqli tenglamalar sistemasi

ï

î

ï

í

ì

=

+

=

+

=

+

3

3

3

2

2

2

1

1

1

c

y

b

x

a

c

y

b

x

a

c

y

b

x

a

0

3

3

3

2

2

2

1

1

1

=

=

D

c

b

a

c

b

a

c

b

a

bo`lganda va uning hech qaysi ikkita tenglamasi o`zaro zid bo`lmasa,

birgalikda bo`ladi.

Masalan. Ushbu

ï

î

ï

í

ì

=

+

=

+

=

-

3

4

9

3

6

3

2

y

x

y

x

y

x

 tenglamalar sistemasini yeching.

Yechish:

0

27

72

6

72

27

6

3

4

1

9

1

3

6

3

2

=

+

-

-

+

-

=

-

=

D

J:Tenglamalar sistemasi birgalikda.

5. Uch noma`lumli uchta chiziqli tenglamalar sistemasi

ï

î

ï

í

ì

=

+

+

=

+

+

=

+

+

3

33

32

31

2

23

22

21

1

13

12

11

,

,

b

z

a

y

a

x

a

b

z

a

y

a

x

a

b

z

a

y

a

x

a

ning bоsh dеtеrminanti

0

33

32

31

23

22

21

13

12

11

¹

=

D

a

a

a

a

a

a

a

a

a

bo’lganda yagоna yеchimga ega bo’lib, bu yеchim Kramеr fоrmulalari bilan

hisоblanadi:

,

,

,

D

D

=

D

D

=

D

D

=

z

y

x

z

y

x

bunda

20

.

,

,

3

32

31

2

22

21

1

12

11

33

3

31

23

2

21

13

1

11

33

32

3

23

22

2

13

12

1

b

a

a

b

a

a

b

a

a

a

b

a

a

b

a

a

b

a

a

a

b

a

a

b

a

a

b

z

y

x

=

D

=

D

=

D

Masalan: Ushbu

ï

î

ï

í

ì

-

=

+

-

=

-

+

-

=

+

-

6

2

3

2

,

8

2

3

,

4

2

z

y

x

z

y

x

z

y

x

chiziqli tеnglamalar sistеmasini yеching.

Yechilishi: asosiy va yordamchi dеtеrminantlarni tоpamiz:

.

4

)

5

(

1

)]

2

(

3

2

)

1

(

)

3

(

1

1

2

2

[

)

1

(

)

2

(

2

1

)

3

(

3

2

2

1

2

3

2

1

2

3

1

2

1

=

-

-

-

=

-

×

×

+

-

×

-

×

+

×

×

-

-

×

-

×

+

×

-

×

+

×

×

=

-

-

-

=

D

Dеtеrminant

0

4

¹

=

D

 bo’lgani uchun sistеma yagоna yеchimga ega va Kramеr

fоrmulasini qo’llab, uni tоpamiz:

;

4

)

56

(

52

)]

2

(

8

2

)

1

(

)

3

(

)

4

(

1

2

)

6

[(

)

1

(

)

2

(

)

6

(

1

)

3

(

8

2

2

4

2

3

6

1

2

8

1

2

4

=

-

-

-

=

=

-

×

×

+

-

×

-

×

-

+

×

×

-

-

-

×

-

×

-

+

×

-

×

+

×

×

-

=

-

-

-

-

-

=

D

x

;

8

)

2

(

6

)]

4

(

3

2

)

1

(

)

6

(

1

1

8

2

[

)

1

(

)

4

(

2

1

)

6

(

3

2

8

1

2

6

2

1

8

3

1

4

1

=

-

-

=

=

-

×

×

+

-

×

-

×

+

×

×

-

-

×

-

×

+

×

-

×

+

×

×

=

-

-

-

=

D

y

.

4

)

4

(

8

)]

2

(

3

)

6

(

8

)

3

(

1

)

4

(

2

2

[

8

)

2

(

2

)

4

(

)

3

(

3

)

6

(

2

1

6

3

2

8

2

3

4

2

1

-

=

-

-

-

=

=

-

×

×

-

+

×

-

×

+

-

×

×

-

×

-

×

+

-

×

-

×

+

-

×

×

=

-

-

-

-

=

D

z

1

4

4

,

2

4

8

,

1

4

4

-

=

-

=

D

D

=

=

=

D

D

=

=

=

D

D

=

z

y

x

z

y

x

J:

.

1

,

2

,

1

-

=

=

=

z

y

x

7. Gauss usuli bilan tenglamalar sistemasini yechish.

Masalan: Ushbu

21

ï

ï

î

ï

ï

í

ì

-

=

+

+

+

=

+

+

+

-

=

+

+

+

=

+

+

+

3

2

3

2

,

2

5

11

3

2

,

3

4

3

,

1

2

5

t

z

y

x

t

z

y

x

t

z

y

x

t

z

y

x

chiziqli tеnglamalar sistеmasini Gauss usuli bilan yеching.

Yechish:Ikkinchi, uchinchi, to’rtinchi tеnlamalardan х larni yo’qоtamiz. Buning uchun

birinchi tеnglamani kеtma-kеt -1, -2, -2  ga ko’paytiramiz va mоs ravishda ikkinchi,

uchinchi, to’rtinchi tеnglamalar bilan qo’shamiz. Natijada ushbu sistеmaga ega

bo’lamiz:

ï

ï

î

ï

ï

í

ì

-

=

-

-

-

=

+

+

=

-

=

+

+

+

,

5

2

7

,

0

,

4

2

2

,

1

2

5

t

z

y

t

z

y

t

z

t

z

y

x

yoki

ï

ï

î

ï

ï

í

ì

=

-

=

+

+

=

+

+

=

+

+

+

.

2

,

5

2

7

,

0

,

1

2

5

t

z

t

z

y

t

z

y

t

z

y

x

Uchinchi tеnglamadan ikkinchi tеnglamani ayiramiz:

ï

ï

î

ï

ï

í

ì

=

-

=

+

=

+

+

=

+

+

+

,

2

,

5

6

,

0

,

1

2

5

t

z

t

z

t

z

y

t

z

y

x

so’ngra to’rtinchi tеnglamani -6 ga ko’paytirib, uchinchi tеnglamaga qo’shsak,

uchburchakli sistеma hоsil bo’ladi:

ï

ï

î

ï

ï

í

ì

-

=

=

-

=

+

+

=

+

+

+

.

7

7

,

2

,

0

,

1

2

5

t

t

z

t

z

y

t

z

y

x

Bundan,

22

.

2

2

5

1

,

0

,

1

2

,

1

-

=

-

-

-

=

=

-

-

=

=

+

=

-

=

t

z

y

x

t

z

y

t

z

t

J:

1

,

1

,

0

,

2

-

=

=

=

-

=

t

z

y

x

.

6. n ta nоma’lumli n ta chiziqli tеnglamalar sistеmasini

ï

ï

î

ï

ï

í

ì

=

+

+

+

=

+

+

+

=

+

+

+

n

n

nn

n

n

n

n

n

n

b

x

a

x

a

x

a

b

x

a

x

a

x

a

b

x

a

x

a

x

a

...

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

,

...

,

...

2

2

1

1

2

2

2

22

1

21

1

1

2

12

1

11

matritsa ko’rinishda

B

AX

=

kabi yozish mumkin, bunda

.

...

,

...

,

.

.

.

.

.

.

.

.

.

.

.

.

.

.

.

2

1

2

1

2

1

2

22

21

1

12

11

÷÷

÷

÷

÷

ø

ö

çç

ç

ç

ç

è

æ

=

÷÷

÷

÷

÷

ø

ö

çç

ç

ç

ç

è

æ

=

÷÷

÷

÷

÷

ø

ö

çç

ç

ç

ç

è

æ

=

n

n

nn

n

n

n

n

b

b

b

B

x

x

x

X

a

a

a

a

a

a

a

a

a

A

Agar A maхsusmas matritsa, ya’ni

0

det

¹

A

 bo’lsa, u hоlda bu sistеmaning matritsa

shaklidagi yеchimi ushbu ko’rinishga ega bo’ladi:

.

1

B

A

X

-

=

E

A

A

AA

=

=

-

-

1

1

 ekanini tеkshirish mumkin.

Masalan: Tenglamalar sistemasini matrisa usuli yordamida yechini.

ï

î

ï

í

ì

=

+

-

-

=

-

+

-

=

+

-

5

6

4

5

9

4

2

5

3

2

z

y

x

z

y

x

z

y

x

Yechish. Tenglamalar sistemasi yordamida A matritsani tuzamiz

23

÷

÷

÷

ø

ö

ç

ç

ç

è

æ

-

-

=

÷

÷

÷

ø

ö

ç

ç

ç

è

æ

=

÷

÷

÷

ø

ö

ç

ç

ç

è

æ

-

-

-

=

5

9

5

,

,

6

4

5

4

2

1

1

3

2

B

z

y

x

X

A

Ushbu matritsaning determinantini hisoblaymiz

;

56

18

32

10

4

60

24

6

4

5

4

2

1

1

3

2

=

+

-

-

-

+

=

-

-

-

=

D

Endi matritsaning algebraik to`ldiruvchilarini topamiz

7

5

12

6

5

1

2

)

1

(

14

)

4

18

(

6

4

1

3

)

1

(

14

10

4

4

5

2

1

)

1

(

26

)

20

6

(

6

5

4

1

)

1

(

4

16

12

6

4

4

2

)

1

(

4

22

3

21

4

13

3

12

2

11

=

-

=

-

=

=

+

-

-

=

-

-

-

=

-

=

-

-

=

-

-

=

-

=

+

-

=

-

-

=

-

=

-

=

-

-

-

=

A

A

A

A

A

7

3

4

2

1

3

2

)

1

(

9

)

1

8

(

4

1

1

2

)

1

(

10

2

12

4

2

1

3

)

1

(

7

)

15

8

(

4

5

3

2

)

1

(

6

33

5

32

4

31

5

23

=

+

=

-

-

=

=

-

-

-

=

-

-

=

=

-

=

-

-

-

=

-

=

+

-

-

=

-

-

-

=

A

A

A

A

Teskari matritsani tuzamiz

÷

÷

÷

ø

ö

ç

ç

ç

è

æ

-

-

-

-

=

-

7

7

14

9

7

26

10

14

4

56

1

1

A

B

A

X

1

-

=

 formulaga asosan noma`lumlarni topamiz

3

;

2

;

1

,

3

2

1

56

168

56

112

56

56

168

112

56

56

1

35

63

70

45

63

130

50

126

20

5

7

)

9

(

)

7

(

)

5

(

14

5

9

)

9

(

7

)

5

(

26

5

10

)

9

(

14

)

5

(

4

5

9

5

7

7

14

9

7

26

10

14

4

56

1

=

=

-

=

÷

÷

÷

ø

ö

ç

ç

ç

è

æ-

=

÷

÷

÷

÷

÷

÷

ø

ö

ç

ç

ç

ç

ç

ç

è

æ-

=

÷

÷

÷

ø

ö

ç

ç

ç

è

æ-

=

÷

÷

÷

ø

ö

ç

ç

ç

è

æ

+

+

+

-

+

-

=

÷

÷

÷

ø

ö

ç

ç

ç

è

æ

×

+

-

×

-

+

-

×

-

×

+

-

×

+

-

×

-

×

+

-

×

+

-

×

-

=

÷

÷

÷

ø

ö

ç

ç

ç

è

æ

-

-

÷

÷

÷

ø

ö

ç

ç

ç

è

æ

-

-

-

-

=

÷

÷

÷

ø

ö

ç

ç

ç

è

æ

=

z

y

x

z

y

x

X

J: (-1;2;3)

24

Misollar.Tenglamalar sistemasini yeching:

1.

î

í

ì

=

+

=

+

4

2

3

3

2

y

x

y

x

2.

î

í

ì

=

-

=

-

1

2

6

2

3

x

x

y

x

3.

î

í

ì

=

+

=

+

2

2

4

1

2

y

x

y

x

4.

î

í

ì

=

-

=

-

2

2

1

3

y

ax

y

ax

5.

(

)

(

)

î

í

ì

¹

=

-

-

=

-

n

m

n

y

x

n

m

ny

mx

2

2

2

6.

î

í

ì

=

+

+

=

+

+

0

3

2

5

0

2

2

3

z

y

x

z

y

x

7.

ï

î

ï

í

ì

=

-

=

+

=

-

5

5

4

2

6

3

2

y

x

y

x

y

x

8.

ï

î

ï

í

ì

=

+

+

=

+

+

=

-

-

16

2

3

4

14

3

2

0

5

z

y

x

z

y

x

z

y

x

9.

ï

î

ï

í

ì

=

-

-

=

+

-

=

-

+

0

5

0

6

0

7

z

y

x

z

y

x

z

y

x

10.

ï

î

ï

í

ì

=

+

-

=

+

+

=

-

+

0

2

3

2

2

8

4

3

z

y

x

z

y

x

z

y

x

Tenglamalar sistemasini Gauss usuli bilan yeching.

11.

ï

ï

î

ï

ï

í

ì

=

+

+

=

+

+

=

+

+

=

+

+

23

5

11

4

15

3

8

2

t

y

x

t

z

x

t

z

y

z

y

x

  12.

ï

ï

î

ï

ï

í

ì

=

+

-

=

+

+

-

=

-

-

=

+

-

+

6

2

3

2

16

3

6

2

6

2

3

z

y

x

t

z

y

t

y

x

t

z

y

x

Tenglamalar sistemasini matritsa usuli bilan yeching.

13.

ï

î

ï

í

ì

=

+

+

=

+

+

=

+

-

3

2

5

6

4

2

12

3

z

y

x

z

y

x

z

y

x

 14.

ï

î

ï

í

ì

=

+

-

=

+

-

=

+

+

0

10

11

4

0

4

3

2

0

z

y

x

z

y

x

z

y

x

15.

ï

î

ï

í

ì

=

+

-

+

=

+

-

+

=

+

-

0

4

3

4

0

5

4

5

2

3

2

z

y

x

z

y

x

z

y

x

 17.

ï

î

ï

í

ì

=

+

-

=

+

-

=

+

-

2

5

3

3

4

2

1

3

4

2

z

y

x

z

y

x

z

y

x

 18.

ï

î

ï

í

ì

=

+

-

-

=

-

+

=

+

-

1

2

2

2

2

3

2

2

z

y

x

z

y

x

z

y

x

20.

ï

î

ï

í

ì

=

+

+

=

-

-

=

+

+

6

4

3

1

2

5

3

2

z

y

x

z

y

x

z

y

x

 21.

ï

î

ï

í

ì

=

+

+

=

+

-

=

+

+

17

10

2

5

4

4

9

2

4

3

z

y

x

z

y

x

z

y

x

 22.

ï

î

ï

í

ì

=

+

-

=

+

+

=

+

-

3

2

2

3

3

0

3

2

z

y

x

z

y

x

z

y

x